Optimal. Leaf size=94 \[ -\frac{2 b \sqrt{a+b \sin (c+d x)}}{d}-\frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{d}+\frac{(a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{d} \]
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Rubi [A] time = 0.16744, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2668, 704, 827, 1166, 206} \[ -\frac{2 b \sqrt{a+b \sin (c+d x)}}{d}-\frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{d}+\frac{(a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 2668
Rule 704
Rule 827
Rule 1166
Rule 206
Rubi steps
\begin{align*} \int \sec (c+d x) (a+b \sin (c+d x))^{3/2} \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{(a+x)^{3/2}}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{2 b \sqrt{a+b \sin (c+d x)}}{d}-\frac{b \operatorname{Subst}\left (\int \frac{-a^2-b^2-2 a x}{\sqrt{a+x} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac{2 b \sqrt{a+b \sin (c+d x)}}{d}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{a^2-b^2-2 a x^2}{-a^2+b^2+2 a x^2-x^4} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{d}\\ &=-\frac{2 b \sqrt{a+b \sin (c+d x)}}{d}-\frac{(a-b)^2 \operatorname{Subst}\left (\int \frac{1}{a-b-x^2} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{d}+\frac{(a+b)^2 \operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\sqrt{a+b \sin (c+d x)}\right )}{d}\\ &=-\frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )}{d}+\frac{(a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{d}-\frac{2 b \sqrt{a+b \sin (c+d x)}}{d}\\ \end{align*}
Mathematica [A] time = 0.0962952, size = 89, normalized size = 0.95 \[ \frac{-2 b \sqrt{a+b \sin (c+d x)}+(a-b)^{3/2} \left (-\tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a-b}}\right )\right )+(a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin (c+d x)}}{\sqrt{a+b}}\right )}{d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.378, size = 218, normalized size = 2.3 \begin{align*} -2\,{\frac{b\sqrt{a+b\sin \left ( dx+c \right ) }}{d}}+{\frac{{a}^{2}}{d}\arctan \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}-2\,{\frac{ab}{d\sqrt{-a+b}}\arctan \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{-a+b}}} \right ) }+{\frac{{b}^{2}}{d}\arctan \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}+{\frac{{a}^{2}}{d}{\it Artanh} \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{a+b}}}} \right ){\frac{1}{\sqrt{a+b}}}}+2\,{\frac{ab}{d\sqrt{a+b}}{\it Artanh} \left ({\frac{\sqrt{a+b\sin \left ( dx+c \right ) }}{\sqrt{a+b}}} \right ) }+{\frac{{b}^{2}}{d}{\it Artanh} \left ({\sqrt{a+b\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{a+b}}}} \right ){\frac{1}{\sqrt{a+b}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (d x + c\right ) \sin \left (d x + c\right ) + a \sec \left (d x + c\right )\right )} \sqrt{b \sin \left (d x + c\right ) + a}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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